Given a Dirichlet character $\chi$ modulo $8$ we consider the map $\mu(x)=x/2$ if $x$ is even and $\mu(x)=(3x+\chi(x))/2$ otherwise. (The corresponding map for $\chi$ the trivial Dirichlet character modulo $2$ is therefore the map of the Collatz-conjecture.)
The two Dirichlet characters modulo $8$ have seemingly very different orbits under iteration:
For $\chi(1,3,5,7)=(1,-1,-1,1)$ the behaviour seems to be as for the Collatz-map: Every orbit starting at $n<10^6$ ends in the period $(1,2)$.
For $\chi(1,3,5,7)=(1,1,-1,-1)$, there are seemingly plenty of diverging orbits and at least four periodic orbits given by $$(1,2)$$ $$(5,7,10)$$ $$(13,19,29,43,65,98,49,74,37,55,82,41,62,31,46,23,34,17,26)$$ $$(53,79,118,59,89,137,67,101,151,226,170,85,127,190,95),142,71,106)$$ Seemingly diverging orbits start at $51$ and $83$ (there are probably many others). Seemingly only finitely many odd numbers have preperiodic orbits.
These different behaviours are not completely unexpected:
$\chi(1,3,5,7)=(1,-1,-1,1)$ corresponds roughly to multiplication by $3/4,\leq 3/8,3/2,3/4$ on odd classes modulo $8$. Since the product of these four numbers is $\leq 81/128<1$, the associated map should be roughly contracting.
$\chi(1,3,5,7)=(1,1,-1,-1)$ on the other hand leads roughly to multiplications by $3/4,3/2,3/2,3/2$ with product $81/32>1$ indicating an expanding behaviour under iteration. (The situation is even slightly worse: An analysis of the underlying non-deterministic automaton with states $1,3,5,7$ modulo $8$ shows that the frequency in orbits of odd numbers congruent to $3$ or $7$ modulo $8$ should roughly be 50 percent larger than the frequency of odd numbers congruent to $1$ or $5$.)
Any ideas for rigorous proofs?
(Remark: The corresponding map for the unique Dirichlet character modulo $4$ given by $\chi(1,3)=(1,-1)$ is easy to analyze.)
